Subtree

Question

  1. You have two every large binary trees: T1,
  2. with millions of nodes, and T2, with hundreds of nodes.
  3. Create an algorithm to decide if T2 is a subtree of T1.
  5. Example
  6. T2 is a subtree of T1 in the following case:
  7.        1                3
  8.       / \              /
  9. T1 = 2   3      T2 =  4
  10.         /
  11.        4
  12. T2 isn't a subtree of T1 in the following case:
  13.        1               3
  14.       / \               \
  15. T1 = 2   3       T2 =    4
  16.         /
  17.        4
  18. Note
  19. A tree T2 is a subtree of T1 if there exists a node n in T1 such that
  20. the subtree of n is identical to T2.
  21. That is, if you cut off the tree at node n,
  22. the two trees would be identical.

题解

判断 T2是否是 T1的子树，首先应该在 T1中找到 T2的根节点，找到根节点后两棵子树必须完全相同。所以整个思路分为两步走：找根节点，判断两棵树是否全等。咋看起来极其简单，但实际实现中还是比较精妙的，尤其是递归的先后顺序及条件与条件或的处理。

Java

  1. /**
  2.  * Definition of TreeNode:
  3.  * public class TreeNode {
  4.  *     public int val;
  5.  *     public TreeNode left, right;
  6.  *     public TreeNode(int val) {
  7.  *         this.val = val;
  8.  *         this.left = this.right = null;
  9.  *     }
  10.  * }
  11.  */
  12. public class Solution {
  13.     /**
  14.      * @param T1, T2: The roots of binary tree.
  15.      * @return: True if T2 is a subtree of T1, or false.
  16.      */
  17.     public boolean isSubtree(TreeNode T1, TreeNode T2) {
  18.         if (T2 == null) return true;
  19.         if (T1 == null) return false;
  20.         return identical(T1, T2) || isSubtree(T1.left, T2) || isSubtree(T1.right, T2);
  21.     }
  23.     private boolean identical(TreeNode T1, TreeNode T2) {
  24.         if (T1 == null && T2 == null) return true;
  25.         if (T1 == null || T2 == null) return false;
  26.         if (T1.val != T2.val) return false;
  27.         return identical(T1.left, T2.left) && identical(T1.right, T2.right);
  28.     }
  29. }

源码分析

这道题的异常处理相对 trick 一点，需要理解 null 对子树的含义。另外需要先调用identical再递归调用isSubtree判断左右子树的情况。方法identical中调用.val前需要判断是否为 null, 而后递归调用判断左右子树是否 identical。

复杂度分析

identical 的调用，时间复杂度近似 O(n), 查根节点的时间复杂度随机，平均为 O(m), 故总的时间复杂度可近似为 O(mn).

Reference